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Help w/ a gravity problem

  1. Oct 29, 2004 #1
    Hello, i am a student taking a physics course online since my school doesn't offer it. I have come across one problem that i nor any science teacher I have access to can help me with. If you would please explain it to me then i would be more than grateful. Thank you

    The problem reads...
    "One of the moons of Jupiter is Callisto. It has a mean distance of 1.883 x 10^6 kilometers from Jupiter and has a period of 16.7 days. What is the mass of Jupiter?"

    Can anyone please explain why the professor set up the equation as 4pie^2(R)^3 \ Gt^2 ???

    Thank you
     
  2. jcsd
  3. Oct 29, 2004 #2

    arildno

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    Welcome to PF!
    He's using Kepler's 3.law, if I'm not mistaken.
    (LONG derivation..)
     
  4. Oct 29, 2004 #3
    not so hard actually;
    consider newton's law of gravity, F = GmM/r^2. This law serves as our centripetal force:
    mv^2/r = GmM/r^2. Simplify this and use v = r*(omega)=r*2*pie/T and you'll get rid of m and get the exact equation for M as you have.
     
  5. Oct 29, 2004 #4

    BobG

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    As arildno said, that equation is practically just Kepler's third law, as explained by Newton (i.e. - the equation explains why Kepler's third law is true). The equation's rearranged to find the gravitational constant for Jupiter with one small variation.

    If you divide the universal gravitational constant (G) out of Jupiter's gravitational constant, you'll get the mass of Jupiter. That's why the G is in the denominator of your equation.
     
  6. Oct 29, 2004 #5

    arildno

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    BLAARGH!
    I'm so used to derive this to gain the relation in terms of the semi-major axis that such elegant arguments as yours are overlooked..:redface:
     
  7. Oct 29, 2004 #6
    don't make it too hard on yourself :)
     
  8. Oct 29, 2004 #7
    alright... thank you all very much. I guess i just got a little overwhelmed when he substituted all those equations in. Thanks again! :biggrin:
     
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